The solution of the equation ${\cos ^2}x - 2\cos x = $ $4\sin x - \sin 2x,$ $\,(0 \le x \le \pi )$ is

  • A

    $\pi - {\cot ^{ - 1}}\left( {\frac{1}{2}} \right)$

  • B

    $\pi - {\tan ^{ - 1}}(2)$

  • C

    $\pi + {\tan ^{ - 1}}\left( { - \frac{1}{2}} \right)$

  • D

    None of these

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The value of $\theta $ lying between $0$ and $\pi /2$ and satisfying the equation

$\left| {\,\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\cos }^2}\theta }&{4\sin 4\theta }\\{{{\sin }^2}\theta }&{1 + {{\cos }^2}\theta }&{4\sin 4\theta }\\{{{\sin }^2}\theta }&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$

  • [IIT 1988]

$2{\sin ^2}x + {\sin ^2}2x = 2,\, - \pi < x < \pi ,$ then $x = $

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